Design of a Doubly Reinforced Beam to Eurocode 2 – Worked Example

This article presents a worked example on the design of doubly reinforced concrete beam to EN 1992-1-2004 (Eurocode 2)

The image below shows the Plan view of a reinforced concrete structure, use the data given below to design Beam 2

 

 

Design Data:

Variable load on slab =   5KN/m²

Finishes  =   1.5KN/m²

Unit weight of concrete   =   24KN/m³

Compressive strength of concrete (fck)   =   30N/mm²

Characteristic Strength of main reinforcement (fyk)  =  500N/mm²

Characteristic Strength of Shear reinforcement (fyv) =  500N/mm²

 

To design Beam 2, we will first distribute the slab loads on the beam and then analyze the beam to compute its internal forces. As shown in fig (2) below the tributary area of slab loads on beam 2 is a trapezium, hence we compute and distribute the slap loads on it as follows:

However, to understand the process of distributing slab load to beams in detail, read, “Distribution of Slab loads to beams.”

fig (2)

Analysis

Permanent Load

Characteristic Self-weight of slab = 0.2 x 24 = 4.8KN/m²

Partition Load on Slab = 1.5KN/m²

Characteristic Permanent Load on Slab   = 4.8 + 1.5 = 6.38KN/m²

Area of Slab load Supported by Beam 2   = (0.5 x (8 + 3) x 2.5)/8 = 1.73m²

Permanent Load of Slab on Beam 2 = 1.72 x 6.38 = 10.97KN/m

Self-weight of beam = 0.23 x 0.45 x 24    = 2.48KN/m

Total Permanent Load on beam = 10.97 +2.48 = 13.5 KN/m

Variable Load    

Variable load on slab = 5KN/m²

Area of Slab load Supported by Beam 2   = (0.5 x (8 + 3) x 2.5)/8   =   1.73m²

Characteristic Variable Load of Slab on Beam 2    = 1.258 x 5   = 8.59KN/m

Total Design Load

Ultimate Load acting on beam 2 = 1.35(13.5) + 1.5(8.59) = 31.11KN/m

 

Computation of Internal Forces:

The beam is assumed to be simply supported for ease of analysis.

M = w x L²/8 =   31 x 8²/8 = 248KNm

V   = W x L/2 =    31 x 8/2   = 124KN

 

Design

flexural strength design

1. Calculate the effective depth

Assumptions

Cover  =   25mm

Main reinforcement diameter = 16mm

Diameter of links =  10mm

Effective depth = h-c-ᴓ-ᴓ/2

         = 450-25-10-16/2

        = 407mm

 

2) Check whether section is to be designed as singly or doubly reinforced beam

$
K\,=\,\,\frac{M}{bd^2f_{ck}}
$

$
K\,=\,\,\frac{248×10^6}{225×450^2×30}
$

 

= 0.22

Since K (0.22) > K’ (0.168); design as doubly reinforced.

3) Calculate the lever arm (Z)

$
Z\,\,=\,\,d\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{K’}{1.134}} \right)
$

 

$
Z\,\,=\,\,407\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{0.22}{1.134}} \right)
$

Since 333.4 < 0.95d (386.7): use Z = 333.4

4. Calculate the area of compression steel required.

 

$
A_{sc\,\,=\,\,\frac{M_{Ed\,\,}-\,\,0.168f_{ck}bd^2}{0.87f_{yk}\left( d-d’ \right)}}
$

$
A_{sc\,\,=\,\,\frac{248×10^6-\,\,0.168 x 30 x 225 x 407^2}{0.87 x 500\left( 407-43 \right)}}
$

Asc = 379.9mm²

Provide 2Y16 (399.9mm²)

5. Calculate the area of tensile steel

$
A_{st}\,\,=\,\,\frac{K_{bal}f_{ck}bd^2}{0.87f_{yk}Z}+A_{sc}
$

$
A_{st}\,\,=\,\,\frac{0.168 x 30 x 225 x 407^2}{0.87 x 500 x 333.4}+379.9
$

Ast = 1675.2mm²

Provide 9Y16 (1799mm²)

 

 

Shear Strength Design

1. Check whether the concrete section can resist the shear force without shear reinforcement

 

V_Rdc = (0.12K(100_ρLf_ck)^1/3 + K1σ_cp) b_wd

 

K     =  (1 +√200/d) =  1 + (200/407)^0.5 = 1.7

ρL = Asl/b_wd = 1799/225 x 407 = 0.018

V_Rdc = (0.12 x 1.7 (100 x 0.018 x 30) ^1/3) 225 x 407

V_Rdc = 78.5KN

Since V_Rdc (78.5KN) is less than V_Ed (124KN) then shear reinforcement has to be designed for.

 

2. Calculate the shear resistance using ϴ  = 22

V_Rdmax(22) = 0.124b_wd(1 – f_ck/250)f_ck

= 0.124 x 225 x 407 (1 – 30/250)30

= 299779.9N

= 299.8KN

Shear resistance of links at ϴ  = 22 is adequate

 

3. Calculate Asw/s at ϴ = 22

 

$
\frac{A_{sw}}{s}\,\,\,\,=\,\,\frac{V_{Ed}}{0.78xf_{yk}d\cot 22}
$

$
\frac{A_{sw}}{s}\,\,\,\,=\,\,\frac{124×10^3}{0.78x407x500\cot 22}
$

$\frac{A_{sw}}{s}\,\,\,\,$ = 0.3

4. Check whether $\frac{A_{sw}}{s}\,\,\,\,$ satisfy the minimum requirement specified by the code.

$$
\frac{A_{sw\min}}{s}\,\,\,\,=\,\,\frac{0.08x\sqrt{f_{ck}}xb_w}{f_{yk}}
$$

$$
\frac{A_{sw\min}}{s}\,\,\,\,=\,\,\frac{0.08x\sqrt{30}x225}{500} $$

= 0.19

Since A_smin/s (0.19) is less than As/s (0.3), shear reinforcement is greater than the required minimum area of reinforcement.

 

5) Calculate shear link reinforcement spacing requirement

Assume two-legged shear reinforcement of 10mm is to be used.

Area of 10mm shear reinforcement = 78.58mm2

Area of two-legged 10mm links = 2x 78.58   = 157mm2

157/s = 0.3

s = 157/0.3

s = 497.9

6) Check whether maximum spacing requirement is satisfied

Smax = 0.75xd

= 0.75 x 407

= 305mm

Since the maximum spacing (305mm) is less than the calculated spacing of links (497.9mm), the maximum spacing limit governs the design.

provide Y10 @ 300mm spacing.

Read also: Design of a Singly Reinforced Beam to Eurocode 2

Deflection Check

1. Calculate the actual span-effective depth ratio

Span/depth ratio = 8000/407   = 19.7

2. Calculate the limiting Span-effective depth ratio

l/d = K[11 + 1.5√f_ck ρ0/ρ + 3.2√f_ck (ρ^o/ρ – 1)3/2] if ρ ≤ ρ^o

l/d = K[11 + 1.5√fck ρ^o/ρ + 3.2√fck √ρ^o/ρ ] if ρ > ρ^o

ρ = Asrequired/b x d

ρ = 1675.2/225 x 407

= 0.018

ρ^o = 10^-3√fck

ρ^o = 10^-3√30

= 0.005

K = 1 (for simply supported)

Since ρ > ρ^o  = then we will use

l/d = K[11 + 1.5√fck ρ^o/ρ + 3.2√fck √ρ^o/ρ ]

l/d = [11 + 1.5√30 0.005/0.019 + 3.2√30 √0.005/0.019]

= 14.3mm                  

Since actual span-effective depth ratio (19.7) is greater than the limiting span-effective depth ratio (14.3), the beam fails deflection check.

NB: Since the beam fails deflection check, it shall be resized by increasing its depth so as to reduce the span-depth ratio. Another effective solution is to design the beam as encastre beam fixed at both ends; this will reduce its overall deflection as well as its sagging moment which will result to the decrease in the area of tensile steel provided for flexural strength of the member.